Question

我是python的新手，我正在尝试使用元组作为键和嵌套列表作为多个值来创建字典。

列表嵌套在三元组中; [[[Isolation source],[host],[country]]...etc]

以下示例：

value_list = [[['NaN'], ['sponge'], ['Palau']], [['skin'], ['fish'], ['Cuba']], [['claw'], ['crab'], ['Japan: Aomori, Natsudomari peninsula']]....]

键的元组;

key_tuple = ('AB479448', 'AB479449', 'AB602436',...)

因此，我希望输出看起来像这样;

dict = {'AB479448': [NaN, sponge, Palau], 'AB479449': [skin, fish, Cuba], 'AB602436': [claw, crab, Japan: Aomori, Natsudomari peninsula]

我尝试了一些不同的解决方案但不是我可以开展工作......例如字典理解;

dict = { i: value_list for i in key_tuple }

上面给出了这个（使用不同的键但是将相同的值与每个键相关联）;

{'AB479448': [[[NaN, sponge, Palau]]], 'AB479449': [[[NaN, sponge, Palau]]], 'AB602436': [[[NaN, sponge, Palau]]]...etc..}

非常感谢任何指示......谢谢！

Answer 1

您可以使用itertools.chain.from_iterable，itertools.izip（或zip）和词典理解：

>>> from itertools import chain, izip
>>> value_list = [[['NaN'], ['sponge'], ['Palau']], [['skin'], ['fish'], ['Cuba']], [['claw'], ['crab'], ['Japan: Aomori, Natsudomari peninsula']]]
>>> key_tuple = ('AB479448', 'AB479449', 'AB602436')
>>> {k: list(chain.from_iterable(v)) for k, v in izip(key_tuple, value_list)}
{'AB479449': ['skin', 'fish', 'Cuba'],
 'AB479448': ['NaN', 'sponge', 'Palau'],
 'AB602436': ['claw', 'crab', 'Japan: Aomori, Natsudomari peninsula']}

Answer 2

使用zip和iter，您可以按如下方式创建所需的输出字典

value_list = [[['NaN'], ['sponge'], ['Palau']], [['skin'], ['fish'], ['Cuba']], [['claw'], ['crab'], ['Japan: Aomori, Natsudomari peninsula']]]
key_tuple = ('AB479448', 'AB479449', 'AB602436')

dict( (key,[[list(value)]]) for key,value in zip(key_tuple, zip(*(iter(t[0] for v in value_list for t in v),)*3)))

Out[16]: {'AB479448': [[['NaN', 'sponge', 'Palau']]], 'AB479449': [[['skin', 'fish', 'Cuba']]],'AB602436': [[['claw', 'crab', 'Japan: Aomori, Natsudomari peninsula']]]}

如果列表键中所需的元素数量发生变化，您可以用3替换新的元素长度值。

制作它真的很有趣。

Answer 3

以下是使用itertools.chain.from_iterable和字典理解的解决方案：

from itertools import chain
{keys[i]:list(chain.from_iterable(contents)) for i, contents in enumerate(my_list)}

这等于：

from itertools import chain
for i, contents in enumerate(my_list): #get [['skin'], ['fish'], ['Cuba']]
    result[keys[i]] = list(chain.from_iterable(contents))

<强>演示：

>>> from itertools import chain
>>> my_list = [[['NaN'], ['sponge'], ['Palau']], [['skin'], ['fish'], ['Cuba']], [['claw'], ['crab'], ['Japan: Aomori, Natsudomari peninsula']]]
>>> keys = ('AB479448', 'AB479449', 'AB602436')
>>> {keys[i]:list(chain.from_iterable(contents)) for i, contents in enumerate(my_list)}
{'AB479449': ['skin', 'fish', 'Cuba'], 'AB479448': ['NaN', 'sponge', 'Palau'], 'AB602436': ['claw', 'crab', 'Japan: Aomori, Natsudomari peninsula']}
>>>

希望这有帮助！

Python：使用嵌套列表中的多个值从元组键创建一个dict

3 个答案: